A parallel plate capacitor with a surface charge density of 6.8 micro-coulombs/m

Question

A parallel plate capacitor with a surface charge density of 6.8 micro-coulombs/m2 hasa vertical orientation and another infinite sheet with a surface charge density of +3.6 micro-coulombs/m2 lies horizontally beneath it as indicated in the sketch. What is the change in electric potential if one moves a distance of 0.38 meters at an angle of 22 degrees below the horizontal? Answer in 104 volts. positive sheet of capacitor negative sheet of capacitor Infinite sheet Hint: find the change in potential for just the capacitor for this path and then find the change in potential for just the sheet for this path. Add both together.

Explanation / Answer

Change in potential = P1 + P2

= -E1*d*cos theta +E2*d sin theta

= -sigma/e0 * d *cos theta +sigma/2e0*d sin theta

= -6.8e-6/8.85e-12 *0.38* cos 22 degree + 6.8e-6/(2*8.85e-12) *0.38* sin 22 degree

= -216028 V

change = -21.6 *10^4 V

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